93學測數學多選第9題詳解

設 $ABCD$ 為邊長 $1$ 的正方形，$M$ 為 $ABCD$ 的中心。建立坐標系： $A = \left(\frac{1}{2}, \frac{1}{2}, 0\right)$，$B = \left(-\frac{1}{2}, \frac{1}{2}, 0\right)$，$C = \left(-\frac{1}{2}, -\frac{1}{2}, 0\right)$，$D = \left(\frac{1}{2}, -\frac{1}{2}, 0\right)$，$O = (0, 0, h)$，其中 $h^2 + \frac{1}{2} = 4$，$h = \sqrt{\frac{7}{2}}$。 $(1)$ $\overset{\large\rightharpoonup}{OA} + \overset{\large\rightharpoonup}{OB} + \overset{\large\rightharpoonup}{OC} + \overset{\large\rightharpoonup}{OD} = (A+B+C+D) - 4O = (0, 0, -4h) \ne \overset{\large\rightharpoonup}{0}$，錯誤。 $(2)$ $\overset{\large\rightharpoonup}{OA} + \overset{\large\rightharpoonup}{OB} - \overset{\large\rightharpoonup}{OC} - \overset{\large\rightharpoonup}{OD} = A + B - C - D = (0, 2, 0) \ne \overset{\large\rightharpoonup}{0}$，錯誤。 $(3)$ $\overset{\large\rightharpoonup}{OA} - \overset{\large\rightharpoonup}{OB} + \overset{\large\rightharpoonup}{OC} - \overset{\large\rightharpoonup}{OD} = A - B + C - D = (0, 0, 0) = \overset{\large\rightharpoonup}{0}$，正確。 $(4)$ $\overset{\large\rightharpoonup}{OA} \cdot \overset{\large\rightharpoonup}{OB} = \frac{1}{2} \times \left(-\frac{1}{2}\right) + \frac{1}{2} \times \frac{1}{2} + h^2 = h^2$；$\overset{\large\rightharpoonup}{OC} \cdot \overset{\large\rightharpoonup}{OD} = \left(-\frac{1}{2}\right) \times \frac{1}{2} + \left(-\frac{1}{2}\right) \times \left(-\frac{1}{2}\right) + h^2 = h^2$。故 $\overset{\large\rightharpoonup}{OA} \cdot \overset{\large\rightharpoonup}{OB} = \overset{\large\rightharpoonup}{OC} \cdot \overset{\large\rightharpoonup}{OD}$，正確。 $(5)$ $\overset{\large\rightharpoonup}{OA} \cdot \overset{\large\rightharpoonup}{OC} = \frac{1}{2} \times \left(-\frac{1}{2}\right) + \frac{1}{2} \times \left(-\frac{1}{2}\right) + h^2 = h^2 - \frac{1}{2} = \frac{7}{2} - \frac{1}{2} = 3 \ne 2$，錯誤。 故答案為 $(3)(4)$。